Prove |G| = (G : H)|H|. We're given that G is a finite group and H is a subgroup of G. The notation (G : H) means the number of left cosets of H in G.
The book tells me that (G : H) = |G|/|H| since every coset of H contains |H| elements.
I'm just having trouble making sense of this, let alone formulating a proof. So any guidance would be much appreciated!
 A: The left cosets of $H$ in $G$ partition $G$, and each coset has $|H|$ elements (since $H$ is a coset and any two cosets of $H$ have the same cardinality). So since there are $(G : H)$ left cosets of $H$ in $G$, $|G| = (G : H)|H|$.
A: Idea is as follows:
1) Show that $|H|=|gH|$ for all $g\in G$.
2) Show that $G/H$, the set of left cosets on $G$, is a partition on $G$ ($g_1H$ and $g_2H$ are either disjoint or equivalent). So $G$ is the union of all its cosets.
3) Since $G$ is finite, we can write $G/H=\{H,g_1H,\dots, g_nH\}$ (there must be only finitely many cosets).
4) $|G|=|H\cup\bigcup_1^ng_iH|=|H|+\sum_1^n |g_iH|=|H|+\sum_1^n |H|=(n+1)|H|=(G:H)|H|$ (the union is disjoint, so the cardinality of the union is the sum of the cardinalities).
A: I like to present this with equivalence relations. Define a relation on $G$ by
$$
x\sim_H y \qquad\text{if and only if}\qquad x^{-1}y\in H.
$$
The fact that this is an equivalence relation is just the same as the laws that state $H$ is a subgroup:


*

*For all $x\in G$, $x\sim_H x$, because $x^{-1}x=1\in H$;

*If $x\sim_H y$, then $x^{-1}y\in H$, so also $(x^{-1}y)^{-1}=y^{-1}x\in H$, which implies $y\sim_H x$;

*If $x\sim_H y$ and $y\sim_H z$, then $x^{-1}y\in H$ and $y^{-1}z\in H$, so also
$x^{-1}yy^{-1}z=x^{-1}z\in H$, which implies $x\sim_H z$.


Now we compute the equivalence class of $1$:
$$
[1]=\{x\in G:1\sim_H x\}=\{x\in G:1^{-1}x\in H\}=H
$$
Similarly, the equivalence class of $g\in G$ is
$$
[g]=\{x\in G:g\sim_H x\}=\{x\in G:g^{-1}x\in H\}.
$$
So, if $x\in G$ we have $g^{-1}x=h\in H$ and $x=gh\in gH$; conversely, if $x=gh$ for some $h\in H$, we have $g^{-1}x=h$, so that $g\sim_H x$.
Next we show that, for every $g\in G$, there is a bijection $H\to gH$; indeed, the required map is $h\mapsto gh$ (the check it's a bijection is trivial).
Until now we haven't used the information that $G$ is finite. In this case, the equivalence relation $\sim_H$ partitions $G$ in pairwise disjoint equivalence classes, all of which have $|H|$ elements. If $(G:H)$ is the number of equivalence classes (the left cosets), then
$$
|G|=|H|(G:H).
$$

There's no preference for the left cosets; you can repeat the same reasoning with the relation
$$
x\sim'_H y \qquad\text{if and only if}\qquad xy^{-1}\in H.
$$
For this relation the equivalence classes are the right cosets.
In particular, the number of distinct right cosets is the same $(G:H)$.
This can also be proved by the fact that there is a bijection from the set of left cosets to the set of right cosets defined by
$$
gH\mapsto Hg^{-1}.
$$
